英文:
Get a list of keys that match each item of another given list...?
问题
result_list = [key for key, values in my_dict.items() if any(item in values for item in search_list)]
英文:
my_dict = {
'NONE': ['N', 'NE'],
'VERY SLIGHT': ['VSLT', 'VT'],
'SLIGHT': ['SLT', 'ST'],
'FAINT': ['F', 'FT'],
'MEDIUM': ['M', 'MM'],
'STRONG': ['S', 'SG'],
'VERY STRONG': ['VS', 'VG']
}
search_list = ['N', 'SLT', 'MM']
result_list = []
I want to search each item of search_list in values of my_dict and return the matching keys as list.
I need... result_list = ['NONE', 'SLIGHT', 'MEDIUM']
I got it by...
for i in search_list:
for k, v in my_dict.items():
if i in v:
result_list.append(k)
how to get it in more simpler and quickest way to do it?
答案1
得分: 1
将文本到代码的字典翻转为代码到文本的字典（每个程序运行一次），然后只需进行简单快速的字典查找。
英文:
"Flip" your text-to-codes dict to a code-to-text dict (once per your program), and then it's just a simple, fast dict lookup away.
text_to_codes = {
"NONE": ["N", "NE"],
"VERY SLIGHT": ["VSLT", "VT"],
"SLIGHT": ["SLT", "ST"],
"FAINT": ["F", "FT"],
"MEDIUM": ["M", "MM"],
"STRONG": ["S", "SG"],
"VERY STRONG": ["VS", "VG"],
}
code_to_text = {}
for text, codes in text_to_codes.items():
code_to_text.update({code: text for code in codes})
# `code_to_text` will end up looking like
# {'N': 'NONE', 'NE': 'NONE', 'VSLT': 'VERY SLIGHT', 'VT': 'VERY SLIGHT', 'SLT': 'SLIGHT', ...
# at this point.
search_list = ["N", "SLT", "MM"]
result_list = [code_to_text[code] for code in search_list]
print(result_list) # ['NONE', 'SLIGHT', 'MEDIUM']
答案2
得分: 0
嵌套推导是一种选择。
[k for k, v in my_dict.items() for x in search_list if x in v]
英文:
Nested comprehension is one option.
[k for k, v in my_dict.items() for x in search_list if x in v]
答案3
得分: 0
Sure, here's the translated code:
search_list = ['N', 'SLT', 'MM']
find_list = []
for k, v in my_dict.items():
find_list.extend(map(lambda x: (x, k), v))
dict_result = {tup[0]: tup[1:] for tup in find_list}
for i in search_list:
print(dict_result[i])
Please note that I've translated the code as requested, but I recommend using a more meaningful variable name than "dict" to avoid conflicts with the built-in Python dict type.
英文:
search_list = ['N', 'SLT', 'MM']
find_list=[]
for k,v in my_dict.items():
find_list.extend(map(lambda x: (x,k), v))
dict = {tup[0]: tup[1:] for tup in find_list}
for i in search_list:
print(dict[i])
答案4
得分: -1
You could use list comprehension:
search_list = ["N", "SLT", "MM"]
result_list = [k for k, v in my_dict.items() for x in search_list if x in v]
This adds the appropriate keys to the result list.
(Add a key only if one of its items is the abbreviation you want.)
On the other hand, you could flip your dictionary, then just look it up:
code_to_text = {}
for k, v in text_to_codes.items():
code_to_text.update({code: k for code in v})
search_list = ["N", "SLT", "MM"]
result_list = [code_to_text[code] for code in search_list]
Both of these give the same value of result_list:
['NONE', 'SLIGHT', 'MEDIUM']
英文:
You could use list comprehension:
search_list = ["N", "SLT", "MM"]
result_list = [k for k, v in my_dict.items() for x in search_list if x in v]
This adds the appropriate keys to the result list.
(Add a key only if one of its items is the abbreviation you want.)
On the other hand, you could flip your dictionary, then just look it up:
code_to_text = {}
for k, v in text_to_codes.items():
code_to_text.update({code: k for code in v})
search_list = ["N", "SLT", "MM"]
result_list = [code_to_text[code] for code in search_list]
Both of these give the same value of result_list:
['NONE', 'SLIGHT', 'MEDIUM']
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论